Largest Nonnegative Integer in Limit of Exponential Expression

To find the largest nonnegative integer value of $a$ for which the given limit holds, let us analyze the limit:

$\lim_{x \to 1} \left\{ \frac{-ax + \sin(x - 1) + a}{x + \sin(x - 1) - 1} \right\}^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4}$

First, simplify the exponent: For $x > 0$ and $x \neq 1$, we have: $\frac{1 - x}{1 - \sqrt{x}} = \frac{(1 - \sqrt{x})(1 + \sqrt{x})}{1 - \sqrt{x}} = 1 + \sqrt{x}$

As $x \to 1$, the exponent approaches: $\lim_{x \to 1} (1 + \sqrt{x}) = 1 + \sqrt{1} = 2$

Next, let $t = x - 1$. As $x \to 1$, we have $t \to 0$. The base of the expression can be rewritten in terms of $t$ as: $\frac{-a(t + 1) + \sin t + a}{(t + 1) + \sin t - 1} = \frac{\sin t - at}{t + \sin t}$

Dividing the numerator and the denominator by $t$ (for $t \neq 0$): $\frac{\frac{\sin t}{t} - a}{1 + \frac{\sin t}{t}}$

Since $\lim_{t \to 0} \frac{\sin t}{t} = 1$, the limit of the base as $x \to 1$ is: $\lim_{t \to 0} \frac{\frac{\sin t}{t} - a}{1 + \frac{\sin t}{t}} = \frac{1 - a}{1 + 1} = \frac{1 - a}{2}$

For the function $[f(x)]^{g(x)}$ to be well-defined in the real domain in a deleted neighborhood of $x = 1$, the base $f(x)$ must be strictly positive since the exponent $g(x) = 1 + \sqrt{x}$ assumes arbitrary irrational values as $x \to 1$. Thus, we require: $\lim_{x \to 1} f(x) > 0 \implies \frac{1 - a}{2} > 0 \implies a < 1$

Now, evaluating the limit: $\lim_{x \to 1} [f(x)]^{g(x)} = \left(\frac{1 - a}{2}\right)^2$

Equating this to the given value: $\left(\frac{1 - a}{2}\right)^2 = \frac{1}{4}$ $\implies (1 - a)^2 = 1$ $\implies 1 - a = \pm 1$

This gives two possible values:

1. $1 - a = 1 \implies a = 0$
2. $1 - a = -1 \implies a = 2$

Since we must have $a < 1$ for the base to be positive (and the expression to be real-valued near $x = 1$), $a = 2$ is extraneous and rejected.

Thus, the only valid value is $a = 0$.

Since $a = 0$ is a nonnegative integer, the largest nonnegative integer value of $a$ is $0$.